3.99 \(\int \cosh ^2(c+d x) (a+b \tanh ^2(c+d x))^3 \, dx\)

Optimal. Leaf size=78 \[ \frac{b^2 (3 a+2 b) \tanh (c+d x)}{d}+\frac{(a+b)^3 \sinh (c+d x) \cosh (c+d x)}{2 d}+\frac{1}{2} x (a-5 b) (a+b)^2+\frac{b^3 \tanh ^3(c+d x)}{3 d} \]

[Out]

((a - 5*b)*(a + b)^2*x)/2 + ((a + b)^3*Cosh[c + d*x]*Sinh[c + d*x])/(2*d) + (b^2*(3*a + 2*b)*Tanh[c + d*x])/d
+ (b^3*Tanh[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.089681, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3675, 390, 385, 206} \[ \frac{b^2 (3 a+2 b) \tanh (c+d x)}{d}+\frac{(a+b)^3 \sinh (c+d x) \cosh (c+d x)}{2 d}+\frac{1}{2} x (a-5 b) (a+b)^2+\frac{b^3 \tanh ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[c + d*x]^2*(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

((a - 5*b)*(a + b)^2*x)/2 + ((a + b)^3*Cosh[c + d*x]*Sinh[c + d*x])/(2*d) + (b^2*(3*a + 2*b)*Tanh[c + d*x])/d
+ (b^3*Tanh[c + d*x]^3)/(3*d)

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \cosh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^3}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (b^2 (3 a+2 b)+b^3 x^2+\frac{(a-2 b) (a+b)^2+3 b (a+b)^2 x^2}{\left (1-x^2\right )^2}\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{b^2 (3 a+2 b) \tanh (c+d x)}{d}+\frac{b^3 \tanh ^3(c+d x)}{3 d}+\frac{\operatorname{Subst}\left (\int \frac{(a-2 b) (a+b)^2+3 b (a+b)^2 x^2}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{(a+b)^3 \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac{b^2 (3 a+2 b) \tanh (c+d x)}{d}+\frac{b^3 \tanh ^3(c+d x)}{3 d}+\frac{\left ((a-5 b) (a+b)^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=\frac{1}{2} (a-5 b) (a+b)^2 x+\frac{(a+b)^3 \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac{b^2 (3 a+2 b) \tanh (c+d x)}{d}+\frac{b^3 \tanh ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.850207, size = 69, normalized size = 0.88 \[ \frac{4 b^2 \tanh (c+d x) \left (9 a-b \text{sech}^2(c+d x)+7 b\right )+6 (a-5 b) (a+b)^2 (c+d x)+3 (a+b)^3 \sinh (2 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[c + d*x]^2*(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

(6*(a - 5*b)*(a + b)^2*(c + d*x) + 3*(a + b)^3*Sinh[2*(c + d*x)] + 4*b^2*(9*a + 7*b - b*Sech[c + d*x]^2)*Tanh[
c + d*x])/(12*d)

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Maple [B]  time = 0.047, size = 148, normalized size = 1.9 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ({\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +3\,{a}^{2}b \left ( 1/2\,\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) -1/2\,dx-c/2 \right ) +3\,a{b}^{2} \left ( 1/2\,{\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{\cosh \left ( dx+c \right ) }}-3/2\,dx-3/2\,c+3/2\,\tanh \left ( dx+c \right ) \right ) +{b}^{3} \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{2\, \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}-{\frac{5\,dx}{2}}-{\frac{5\,c}{2}}+{\frac{5\,\tanh \left ( dx+c \right ) }{2}}+{\frac{5\, \left ( \tanh \left ( dx+c \right ) \right ) ^{3}}{6}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^3,x)

[Out]

1/d*(a^3*(1/2*cosh(d*x+c)*sinh(d*x+c)+1/2*d*x+1/2*c)+3*a^2*b*(1/2*cosh(d*x+c)*sinh(d*x+c)-1/2*d*x-1/2*c)+3*a*b
^2*(1/2*sinh(d*x+c)^3/cosh(d*x+c)-3/2*d*x-3/2*c+3/2*tanh(d*x+c))+b^3*(1/2*sinh(d*x+c)^5/cosh(d*x+c)^3-5/2*d*x-
5/2*c+5/2*tanh(d*x+c)+5/6*tanh(d*x+c)^3))

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Maxima [B]  time = 1.15486, size = 346, normalized size = 4.44 \begin{align*} \frac{1}{8} \, a^{3}{\left (4 \, x + \frac{e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac{3}{8} \, a^{2} b{\left (4 \, x - \frac{e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac{1}{24} \, b^{3}{\left (\frac{60 \,{\left (d x + c\right )}}{d} + \frac{3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac{121 \, e^{\left (-2 \, d x - 2 \, c\right )} + 201 \, e^{\left (-4 \, d x - 4 \, c\right )} + 147 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )}\right )}}\right )} - \frac{3}{8} \, a b^{2}{\left (\frac{12 \,{\left (d x + c\right )}}{d} + \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac{17 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )}\right )}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

1/8*a^3*(4*x + e^(2*d*x + 2*c)/d - e^(-2*d*x - 2*c)/d) - 3/8*a^2*b*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)
/d) - 1/24*b^3*(60*(d*x + c)/d + 3*e^(-2*d*x - 2*c)/d - (121*e^(-2*d*x - 2*c) + 201*e^(-4*d*x - 4*c) + 147*e^(
-6*d*x - 6*c) + 3)/(d*(e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + 3*e^(-6*d*x - 6*c) + e^(-8*d*x - 8*c)))) - 3/8*
a*b^2*(12*(d*x + c)/d + e^(-2*d*x - 2*c)/d - (17*e^(-2*d*x - 2*c) + 1)/(d*(e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c)
)))

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Fricas [B]  time = 2.0419, size = 903, normalized size = 11.58 \begin{align*} \frac{3 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sinh \left (d x + c\right )^{5} - 4 \,{\left (18 \, a b^{2} + 14 \, b^{3} - 3 \,{\left (a^{3} - 3 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{3} - 12 \,{\left (18 \, a b^{2} + 14 \, b^{3} - 3 \,{\left (a^{3} - 3 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} +{\left (9 \, a^{3} + 27 \, a^{2} b + 99 \, a b^{2} + 65 \, b^{3} + 30 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} - 12 \,{\left (18 \, a b^{2} + 14 \, b^{3} - 3 \,{\left (a^{3} - 3 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) + 3 \,{\left (5 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cosh \left (d x + c\right )^{4} + 2 \, a^{3} + 6 \, a^{2} b + 30 \, a b^{2} + 10 \, b^{3} +{\left (9 \, a^{3} + 27 \, a^{2} b + 99 \, a b^{2} + 65 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{24 \,{\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 3 \, d \cosh \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

1/24*(3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sinh(d*x + c)^5 - 4*(18*a*b^2 + 14*b^3 - 3*(a^3 - 3*a^2*b - 9*a*b^2 -
5*b^3)*d*x)*cosh(d*x + c)^3 - 12*(18*a*b^2 + 14*b^3 - 3*(a^3 - 3*a^2*b - 9*a*b^2 - 5*b^3)*d*x)*cosh(d*x + c)*s
inh(d*x + c)^2 + (9*a^3 + 27*a^2*b + 99*a*b^2 + 65*b^3 + 30*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^2)*s
inh(d*x + c)^3 - 12*(18*a*b^2 + 14*b^3 - 3*(a^3 - 3*a^2*b - 9*a*b^2 - 5*b^3)*d*x)*cosh(d*x + c) + 3*(5*(a^3 +
3*a^2*b + 3*a*b^2 + b^3)*cosh(d*x + c)^4 + 2*a^3 + 6*a^2*b + 30*a*b^2 + 10*b^3 + (9*a^3 + 27*a^2*b + 99*a*b^2
+ 65*b^3)*cosh(d*x + c)^2)*sinh(d*x + c))/(d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c)*sinh(d*x + c)^2 + 3*d*cosh(d*
x + c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)**2*(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Timed out

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Giac [B]  time = 2.06105, size = 360, normalized size = 4.62 \begin{align*} \frac{12 \,{\left (a^{3} - 3 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\right )} d x - 3 \,{\left (2 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 6 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 18 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 10 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 3 \,{\left (a^{3} e^{\left (2 \, d x + 12 \, c\right )} + 3 \, a^{2} b e^{\left (2 \, d x + 12 \, c\right )} + 3 \, a b^{2} e^{\left (2 \, d x + 12 \, c\right )} + b^{3} e^{\left (2 \, d x + 12 \, c\right )}\right )} e^{\left (-10 \, c\right )} - \frac{16 \,{\left (9 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 9 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 18 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 12 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 9 \, a b^{2} + 7 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")

[Out]

1/24*(12*(a^3 - 3*a^2*b - 9*a*b^2 - 5*b^3)*d*x - 3*(2*a^3*e^(2*d*x + 2*c) - 6*a^2*b*e^(2*d*x + 2*c) - 18*a*b^2
*e^(2*d*x + 2*c) - 10*b^3*e^(2*d*x + 2*c) + a^3 + 3*a^2*b + 3*a*b^2 + b^3)*e^(-2*d*x - 2*c) + 3*(a^3*e^(2*d*x
+ 12*c) + 3*a^2*b*e^(2*d*x + 12*c) + 3*a*b^2*e^(2*d*x + 12*c) + b^3*e^(2*d*x + 12*c))*e^(-10*c) - 16*(9*a*b^2*
e^(4*d*x + 4*c) + 9*b^3*e^(4*d*x + 4*c) + 18*a*b^2*e^(2*d*x + 2*c) + 12*b^3*e^(2*d*x + 2*c) + 9*a*b^2 + 7*b^3)
/(e^(2*d*x + 2*c) + 1)^3)/d